KazakhMO 2018

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Source: Kungozhin

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#1 h May 6, 2018, 2:31 PM • 2 Y

Y by Adventure10, Mango247

The trapezium diagonals $ABCD$ ( $AD \parallel BC$ ) intersect at the point $K$ . The points $L$ and $M$ are marked on the line $AD$ such that $A$ lies on the segment $LD$ , $D$ lies on the segment $AM$ , $AL = AK$ and $DM = DK$ . Prove that the lines $CL$ and $BM$ intersect on the bisector of the angle $BKC$ .

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#2 h May 6, 2018, 2:54 PM • 3 Y

Y by A.L.E.X, Adventure10, Mango247

Easy one
Let angle bisectors of $\angle LAC$ and $\angle BDM$ intersect at $T$ .Easy to notice that $T$ is excenter of $\Delta AKD$ hence $KT$ is angle bisector of $\angle AKT$ .Now let angle bisectors of $\angle KBC$ and $\angle KCB$ intersect at $P$ (I mean outside angles) $\angle BCL=\angle ALC=\frac{\angle CAD}{2}=\frac{\angle BCA}{2}$ same for $\angle CBM$ $\implies$ $BPCG$ is cyclic( $G=BM\cap CL$ ) and $\angle PGB=\angle PCB=\angle PKB$ $\implies$ $P,G,K,T$ are collinear.Done.

This post has been edited 2 times. Last edited by georgeado17, May 6, 2018, 2:56 PM
Reason: typo

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#3 h May 6, 2018, 3:03 PM • 2 Y

Y by Adventure10, Mango247

georgeado17 wrote:

Easy one
Let angle bisectors of $\angle LAC$ and $\angle BDM$ intersect at $T$ .Easy to notice that $T$ is excenter of $\Delta AKD$ hence $KT$ is angle bisector of $\angle AKT$ .Now let angle bisectors of $\angle KBC$ and $\angle KCB$ intersect at $P$ (I mean outside angles) $\angle BCL=\angle ALC=\frac{\angle CAD}{2}=\frac{\angle BCA}{2}$ same for $\angle CBM$ $\implies$ $BPCG$ is cyclic( $G=BM\cap CL$ ) and $\angle PGB=\angle PCB=\angle PKB$ $\implies$ $P,G,K,T$ are collinear.Done.

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